A drop of some liquid of volume `0.04 cm^(3)` is placed on the surface of a glass slide. Then, another glass forms a thin layer of area `20 cm^(2)` between the surfaces of the two slides. To separate the slides a force of `16 xx 10^(5)` dyne has to be applied normal to the surfaces. The surface tension of the liquid is (in dyne `cm^(-1))`

Options

A. 60
B. 70
C. 80
D. 90

Explanation

To find the surface tension of the liquid, we can follow these steps:

### Step 1: Write down the given values
- Volume of the liquid (V) = 0.04 cm³
- Area of the layer (A) = 20 cm²
- Force applied (F) = 16 × 10⁵ dyne

### Step 2: Calculate the thickness (length) of the liquid layer
The thickness (l) of the liquid layer can be calculated using the formula:
\[ l = \frac{V}{A} \]

Substituting the values:
\[ l = \frac{0.04 \, \text{cm}^3}{20 \, \text{cm}^2} = 0.002 \, \text{cm} \]

### Step 3: Calculate the radius (r)
Since the thickness of the liquid layer is considered as the diameter, the radius (r) is half of the thickness:
\[ r = \frac{l}{2} = \frac{0.002 \, \text{cm}}{2} = 0.001 \, \text{cm} \]

### Step 4: Relate surface tension (S) to the pressure (P)
The excess pressure (P) due to surface tension is given by:
\[ P = \frac{S}{r} \]
Where:
- S = surface tension
- r = radius

### Step 5: Calculate pressure (P) using force and area
Pressure can also be defined as:
\[ P = \frac{F}{A} \]
Substituting the values:
\[ P = \frac{16 \times 10^5 \, \text{dyne}}{20 \, \text{cm}^2} = 8 \times 10^4 \, \text{dyne/cm}^2 \]

### Step 6: Set the two expressions for pressure equal to each other
From the previous steps, we have:
\[ \frac{S}{r} = \frac{F}{A} \]
Substituting for P:
\[ \frac{S}{0.001 \, \text{cm}} = 8 \times 10^4 \, \text{dyne/cm}^2 \]

### Step 7: Solve for surface tension (S)
Rearranging the equation gives:
\[ S = P \cdot r \]
Substituting the values:
\[ S = (8 \times 10^4 \, \text{dyne/cm}^2) \cdot (0.001 \, \text{cm}) = 80 \, \text{dyne/cm} \]

### Final Answer
The surface tension of the liquid is **80 dyne/cm**.